证明: sin7x+sinx⼀sin5x+sin3x=4cos^2 x-3

2025-01-03 15:31:30
推荐回答(3个)
回答1:

证明
分子
sin(4x+3x)+sin(4x-3x)

=sin4xcos3x+cos4xsin4x+sin4xcos3x-cos4xsin4x
=2sin4xcos3x
分母
sin(4x+x)+sin(4x-x)
=sin4xcosx+cos4xsinx+sin4xcosx-cos4xsinx
=2sin4xcosx
左=(2sin4xcos3x)/(2sin4xcosx)=cos3x/cosx,其中
cos3x
=cos(2x+x)
=cos2xcosx-sin2xsinx
=cos2xcosx-2cosx(sinx)^2
所以cos3x/cosx变为
cos2x-2(sinx)^2
=[2(cosx)^2-1]-2*[1-(cosx)^2]
=4(cosx)^2-3
左=右
证毕

回答2:

sin7x+sinx=2sin(7x+x)cos(7x-x)=2sin8xcosx6x
sin5x+sin3x=2sin(5x+3x)cosx(5x-3x)=2sin8xcos2x
左边=cos6x/cos2x=4cos^3(2x)-3cos2x/cos2x=
4cos^2 x-3

回答3: